2021 H2 Chem P1 Q25 — Redox Solution

The answer is B (1.5).

In \(\text{FeCr}_2\text{O}_4\), Fe is +2, so the chromite ion is \(\text{Cr}_2\text{O}_4^{2-}\), with Cr in the +3 oxidation state. In \(\text{CrO}_4^{2-}\), Cr is +6.

Method 1 (half-equations):
\(\text{Cr}_2\text{O}_4^{2-} + 4\text{O}^{2-} \rightarrow 2\text{CrO}_4^{2-} + 6\text{e}^-\) ×2
\(\text{O}_2 + 4\text{e}^- \rightarrow 2\text{O}^{2-}\) ×3
Overall: \(2\text{Cr}_2\text{O}_4^{2-} + 2\text{O}^{2-} + 3\text{O}_2 \rightarrow 4\text{CrO}_4^{2-}\)
2 mol of chromite ions need 3 mol of \(\text{O}_2\), so 1 mol needs 1.5 mol of \(\text{O}_2\).

Method 2 (electron count): \(\text{Cr}_2\text{O}_4^{2-} \rightarrow 2\text{CrO}_4^{2-}\) involves a change in oxidation number of each Cr from +3 to +6, so one mole of \(\text{Cr}_2\text{O}_4^{2-}\) loses \(2 \times (6-3) = 6\) moles of \(\text{e}^-\).
\(\text{O}_2 + 4\text{e}^- \rightarrow 2\text{O}^{2-}\) : hence \(\frac{6}{4} = 1.5\) moles of \(\text{O}_2\) are required to accept 6 moles of \(\text{e}^-\).

A ✗: 1 mol of \(\text{O}_2\) accepts only 4 mol of \(\text{e}^-\), but 6 mol of \(\text{e}^-\) are lost per mole of chromite ions.

C ✗: 3 mol of \(\text{O}_2\) is the amount needed for 2 mol of chromite ions in the overall equation, not 1 mol.

D ✗: 6 is the number of moles of electrons lost per mole of chromite ions, not the number of moles of \(\text{O}_2\).